# If the sum of the roots of a quadratic equation is 2 and the sum of their cubes is 98, then the equation is ?

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The Brainliest Answer!

a,b

so given

a + b = 2

and

a³ + b³ = 98

⇒ (a + b)³ - 3ab(a + b) = 98

⇒ 2³ - 3ab × 2 = 98

⇒ -6ab = 90

⇒ ab = - 15

so the required equation =

x² - (a + b)x + ab = 0

⇒ x² - 2x - 15 = 0

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Let the roots of quadratic equations are m and n

sum of the roots = 2 : m+n=2

sum of their cubes = 98 : m³+n³=98

(m+n)³ = m³ + n³ + 3mn × (m+n)

⇒2³ = 98 + 3mn(2)

⇒ 8 = 98 + 6 mn

⇒ 6mn = 8-98 =-90

⇒ mn = -90/6

⇒ mn = -15

We know that for a quadratic equation x²+ax+b = 0,

sum of roots (m+n)= -b/1 = 2 ⇒ a = -2

product of roots(mn) = c/1 = -15 ⇒ b = -15

So equation is**x² - 2x -15 = 0**

sum of the roots = 2 : m+n=2

sum of their cubes = 98 : m³+n³=98

(m+n)³ = m³ + n³ + 3mn × (m+n)

⇒2³ = 98 + 3mn(2)

⇒ 8 = 98 + 6 mn

⇒ 6mn = 8-98 =-90

⇒ mn = -90/6

⇒ mn = -15

We know that for a quadratic equation x²+ax+b = 0,

sum of roots (m+n)= -b/1 = 2 ⇒ a = -2

product of roots(mn) = c/1 = -15 ⇒ b = -15

So equation is